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1.Inlet :
A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.
Outlet :
A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.
2. If a pipe can fill a tank in x hours, then :
part filled in 1 hour = 1 / x.
3. I. If a pipe can empty a tank in y hours, then :
part emptied in 1 hour = 1 / y.
II. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes,
then the net part filled in 1 hour = (1 / x - 1 / y).
III. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes,
then the net part emptied in 1 hour = (1 / y - 1 / x).
A. 4 hrs 30 min
B. 7 hrs 30 min
C. 8 hrs 10 min
D. None of these
Answer: B
Explanation:
Net part filled in 1 hour = (1/10) + (1/12)-(1/20) = (8/60) = (2/15).
The tank will be full in 15/2 hrs = 7 hrs 30 min.
A. 14 hours
B. 16 hours
C. 18 hours
D. 21 hours
Answer: D
Explanation:
Work done by the leak in 1 hour= (1/3)-(1/ (7/2))
= (1/3)-(2/7) = (1/21).
The leak will empty .the tank in 21 hours.
A. 17 min
B. 24 min
C. 28 min
D. 39 min
Answer: D
Explanation:
Part filled in 7 min. = 7*((1/36) + (1/45)) = (7/20).
Remaining part= (1-(7/20)) = (13/20).
Net part filled in 1 min. when A, B and C are
opened= (1/36) + (1/45)-(1/30) = (1/60).
Now, (1/60) part is filled in one minute.
(13/20) part is filled in (60*(13/20)) =39 minutes.
A. 5 min
B. 14 min
C. 8 min
D. 19 min
Answer: C
Explanation:
Let B be closed after x min. then,
Part filled by (A+B) in x min. +part filled by A in (18-x) min = 1
Therefore x*((1/24) + (1/32)) + (18-x)*(1/24) =1
=> (7x/96) + ((18-x)/24) =1.
=> 7x +4*(18-x) =96.
Hence, be must be closed after 8 min.
A. 12 min
B. 15 min
C. 25 min
D. 50 min
Answer: A
Explanation:
Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30
Part filled by (A + B) in 1 min = (1/20 + 1/30) = 1/12
Both pipes can fill the tank in 12 minutes.
A. 2
B. 2.5
C. 3
D. 3.5
Explanation:
part filled by (A+B+C) in 1 hour = (1/5 + 1/6 + 1/30)
‹=› 1/3.
All the three pipes together will fill the tank in 3 hours.
A. 5/11
B. 6/11
C. 7/11
D. 8/11
Solution:
Part filled by (A + B + C) in 3 minutes = 3(1/30+1/20+1/10)
= (3×11/60)
= 11/20.
Part filled by C in 3 minutes = 3/10.
Required ratio = (3/10×20/11)
= 6/11.
A. 8 hrs
B. 9 hrs
C. 18 hrs
D. 36 hrs
Solution:
Part filled by (A + B) in 1 hour = (1/5 + 1/20)
= 1/4.
So, A and B together can fill the tank in 4 hours.
Work done by the leak in 1 hour = (1/4 - 2/9)
= 1 / 36.
Therefore, Leak will empty the tank in 36 hrs.
A. 8
B. 15
C. 16
D. 18
Answer: D
Explanation:
Capacity of the tank = (12 * 13.5) litres = 162 litres.
Capacity of each bucket = 9 litres
Number of buckets needed = (162/9) = 18.
A. 5 hrs
B. 7 hrs
C. 6 hrs
D. 8 hrs
Answer: C
Explanation:
Net part filled in 1 hour = (1/2) - (1/3) = 1/6
Cistern will be full in 6 hours.
A. filled in 12 min
B. filled in 8 min
C. emptied in 12 min
D. emptied in 8 min
Answer: D
Explanation:
Rate of waste pipe being more, the tank will be emptied when both the pipes are opened.
Net emptying work done by both in 1 min = (1/8) - (1/16) = 1/16
Now, full tank will be emptied by them in 16 min.
Half full tank will be emptied in 8 min.
A. 3 min and 30 sec.
B. 4 min and 30 sec.
C. 4 min.
D. 4 min 77 sec.
Answer: C
Explanation:
A tank can be filled by a tap in 20 minutes and by another tap in 6O minutesLet B be closed after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (9 — x) min, = 1
x[(1/12) + (1/16)] + (9 - x)(1/12) = 1 or (7x/48) + (9-x)/12 = 1
or7x + 36 — 4x = 48 or x=4.
So, B must be closed after 4 minutes.
A. 10 min
B. 15 min
C. 12 min
D. 20 min
Answer: D
Part filled in 10 min = 10[(1/20) + (1/60)] = 10 * (4/60) = 2/3
Remaining part = (1 - (2/3)) = 1/3
Part filled by second tap in 1 min = 1/60
(1/60) : (1/3) ? 1 : x
Hence, the remaining part will be filled in 20 min.
A. 5260
B. 5846
C. 5760
D. 6970
Answer: C
Explanation:
Work done by the inlet in 1 hour = (1/6) - (1/8) = 1/24
Work done by the inlet in 1 min = (1/24) * (1/60) = 1/1440
Volume of 1/1440 part = 4 liters.
Volume of whole = (1440 * 4) litres = 5760 litres.
A. 5 hrs
B. 12 and 1/2 hrs
C. 6 hrs
D. 7 and 1/2 hrs
Answer: C
Explanation:
A's hours work=1/10, B's 1 hours work = 1/15,
(A+B)s 1 hours work = (1/10) + (1/15) = 5/30 = 1/6
Both the taps can fill the tank in 6 hours.
A. 10 hours 30 min
B. 21 hours
C. 12 hours
D. 24 hours
Answer: B
Explanation:
Work done by the leak in 1 hour = (1/3) - (2/7) = 1/21 .
Leak will mpty the tank in 21 hours.
A. 5
B. 12
C. 10
D. 15 and 2/3
Answer: A
Explanation:
Part filled by (A +B+ c) in 1 min. = (1/20) +(1/15) + (1/12) = 12/60 = 1/5
All the three pipes together will fill the tank in 5 min.
A. 4.5 hrs
B. 6.5 hrs
C. 5 hrs
D. 7.2 hrs
Answer : D
Explanation :
Net part filled in 1 hour = (1/4 - 1/9) = 5/36
The cistern will be filled in 36/5 hrs i.e., 7.2 hrs.